Consider an election with three candidates: A, B and C. Each voter orders the candidates randomly (ABC, ACB, BAC, BCA, CAB or CBA), such that there is a 1/6th chance of voting for each of the options. What is the probability that the party which received the most first preference votes (and hence would have won in a 'first past the post' electoral system) does not win in a preferential electoral system?
Background: In the UK, US, India and many other nations, a 'first past the post' electoral system is used: that is, each voter ticks their preferred candidate, and the candidate with the most votes wins. In elections with more than one candidate, the winning candidate may receive less than 50% of the vote.
In many other nations (including my native Australia), an 'instant runoff' or preferential voting system is used: each voter numbers the candidates in a list of their preference. After the election, the candidates are ordered according to who received the most first preference votes. The least popular party is eliminated, and its votes are redistributed according to the preferences of those who voted for the party. This process is repeated until there are only two parties left, by which stage one will have more than 50% of the vote.
My 'brute force' computer model returns the answer that, around 24% of the time, the different electoral systems return a different result. Any ideas for an analytical solution/proof?